Is $K[X]hookrightarrow K[[X]]$ an epimorphism?Is it true that $(A[[x]]otimes_AmathbbC)cap overlinemathbbQ[[x]] = A[[x]]otimes_AoverlinemathbbQ$?Given a commutative ring $R$ and an epimorphism $R^m to R^n$ is then $m geq n$?epimorphism in the category of commutative ringsIs there a characterization of integral domains in terms of the homomorphisms out of them?Abelian category induced by commutative ringIs I-adic completion a ring epimorphism?Is there a purely category-theoretic description of the total quotient ring?Existence of coproducts and products in categoryProve a certain homomorphism in the category of rings is an epimorphismCanonical homomorphism into scalar extension module through a ring epimorphismEqualizer in the category of rings with 1.
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Is $K[X]hookrightarrow K[[X]]$ an epimorphism?
Is it true that $(A[[x]]otimes_AmathbbC)cap overlinemathbbQ[[x]] = A[[x]]otimes_AoverlinemathbbQ$?Given a commutative ring $R$ and an epimorphism $R^m to R^n$ is then $m geq n$?epimorphism in the category of commutative ringsIs there a characterization of integral domains in terms of the homomorphisms out of them?Abelian category induced by commutative ringIs I-adic completion a ring epimorphism?Is there a purely category-theoretic description of the total quotient ring?Existence of coproducts and products in categoryProve a certain homomorphism in the category of rings is an epimorphismCanonical homomorphism into scalar extension module through a ring epimorphismEqualizer in the category of rings with 1.
$begingroup$
Let $K$ be a field and $X$ an indeterminate.
Is the natural monomorphism $K[X]hookrightarrow K[[X]]$ an epimorphism?
By epimorphism I mean epimorphism in the category of commutative rings.
EDIT. The question can be spelled out as follows: Are there distinct morphisms from $K[[X]]$ to some commutative ring $A$ which coincide on $K[X]$?
abstract-algebra polynomials commutative-algebra formal-power-series
$endgroup$
add a comment |
$begingroup$
Let $K$ be a field and $X$ an indeterminate.
Is the natural monomorphism $K[X]hookrightarrow K[[X]]$ an epimorphism?
By epimorphism I mean epimorphism in the category of commutative rings.
EDIT. The question can be spelled out as follows: Are there distinct morphisms from $K[[X]]$ to some commutative ring $A$ which coincide on $K[X]$?
abstract-algebra polynomials commutative-algebra formal-power-series
$endgroup$
$begingroup$
What is an epimorphism exactly ? $K[[x]]$ is the completion of $K[x]$ for the (non-Archimedian) absolute value $|x^Nf(x)| = 2^-N$ if $f(0) ne 0$
$endgroup$
– reuns
Aug 12 '17 at 12:56
2
$begingroup$
@reuns The inclusion $f:K[X]to K[[X]]$ is epi if for every two ring homomorphisms $g_1,g_2:K[[X]]to R$, if $g_1circ f=g_2circ f$ then $g_1=g_2$.
$endgroup$
– uSir470888
Aug 12 '17 at 13:00
add a comment |
$begingroup$
Let $K$ be a field and $X$ an indeterminate.
Is the natural monomorphism $K[X]hookrightarrow K[[X]]$ an epimorphism?
By epimorphism I mean epimorphism in the category of commutative rings.
EDIT. The question can be spelled out as follows: Are there distinct morphisms from $K[[X]]$ to some commutative ring $A$ which coincide on $K[X]$?
abstract-algebra polynomials commutative-algebra formal-power-series
$endgroup$
Let $K$ be a field and $X$ an indeterminate.
Is the natural monomorphism $K[X]hookrightarrow K[[X]]$ an epimorphism?
By epimorphism I mean epimorphism in the category of commutative rings.
EDIT. The question can be spelled out as follows: Are there distinct morphisms from $K[[X]]$ to some commutative ring $A$ which coincide on $K[X]$?
abstract-algebra polynomials commutative-algebra formal-power-series
abstract-algebra polynomials commutative-algebra formal-power-series
edited Aug 12 '17 at 13:04
Pierre-Yves Gaillard
asked Aug 12 '17 at 11:06
Pierre-Yves GaillardPierre-Yves Gaillard
13.5k23184
13.5k23184
$begingroup$
What is an epimorphism exactly ? $K[[x]]$ is the completion of $K[x]$ for the (non-Archimedian) absolute value $|x^Nf(x)| = 2^-N$ if $f(0) ne 0$
$endgroup$
– reuns
Aug 12 '17 at 12:56
2
$begingroup$
@reuns The inclusion $f:K[X]to K[[X]]$ is epi if for every two ring homomorphisms $g_1,g_2:K[[X]]to R$, if $g_1circ f=g_2circ f$ then $g_1=g_2$.
$endgroup$
– uSir470888
Aug 12 '17 at 13:00
add a comment |
$begingroup$
What is an epimorphism exactly ? $K[[x]]$ is the completion of $K[x]$ for the (non-Archimedian) absolute value $|x^Nf(x)| = 2^-N$ if $f(0) ne 0$
$endgroup$
– reuns
Aug 12 '17 at 12:56
2
$begingroup$
@reuns The inclusion $f:K[X]to K[[X]]$ is epi if for every two ring homomorphisms $g_1,g_2:K[[X]]to R$, if $g_1circ f=g_2circ f$ then $g_1=g_2$.
$endgroup$
– uSir470888
Aug 12 '17 at 13:00
$begingroup$
What is an epimorphism exactly ? $K[[x]]$ is the completion of $K[x]$ for the (non-Archimedian) absolute value $|x^Nf(x)| = 2^-N$ if $f(0) ne 0$
$endgroup$
– reuns
Aug 12 '17 at 12:56
$begingroup$
What is an epimorphism exactly ? $K[[x]]$ is the completion of $K[x]$ for the (non-Archimedian) absolute value $|x^Nf(x)| = 2^-N$ if $f(0) ne 0$
$endgroup$
– reuns
Aug 12 '17 at 12:56
2
2
$begingroup$
@reuns The inclusion $f:K[X]to K[[X]]$ is epi if for every two ring homomorphisms $g_1,g_2:K[[X]]to R$, if $g_1circ f=g_2circ f$ then $g_1=g_2$.
$endgroup$
– uSir470888
Aug 12 '17 at 13:00
$begingroup$
@reuns The inclusion $f:K[X]to K[[X]]$ is epi if for every two ring homomorphisms $g_1,g_2:K[[X]]to R$, if $g_1circ f=g_2circ f$ then $g_1=g_2$.
$endgroup$
– uSir470888
Aug 12 '17 at 13:00
add a comment |
3 Answers
3
active
oldest
votes
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The map from a Noetherian local ring to its completion is faithfully flat, and a faithfully flat epimorphism is an isomorphism. It follows from this that the map from the localization of $K[X]$ at $(X)$ to $K[[X]]$ is not epi.
$endgroup$
1
$begingroup$
See stacks.math.columbia.edu/tag/04VM and stacks.math.columbia.edu/tag/00MC
$endgroup$
– Mariano Suárez-Álvarez
Aug 12 '17 at 14:01
1
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Is there a straightforward way to extract a concrete counterexample from this?
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– tomasz
Aug 12 '17 at 14:08
1
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@Pierre-Yves Gaillard, we have that the inclusion factors as $K[X]to K[X]_(X)to K[[X]]$, so if the inclusion were epi, $K[X]_(X)to K[[X]]$ would have to be as well.
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– Ennar
Aug 12 '17 at 19:26
2
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I don't think that's what they're saying, they're saying "If $fcirc g$ is epi, then so is $f$"
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– Max
Aug 12 '17 at 20:03
1
$begingroup$
Yes, that's what I'm saying, @Max. That $K[X]to K[X]_(X)$ is epimorphism establishes the other direction in "if and only if", but is not needed here.
$endgroup$
– Ennar
Aug 12 '17 at 20:14
|
show 7 more comments
$begingroup$
Let $B$ be a transcendence basis for the field $K((X))$ over $K(X)$ and let $L$ be an algebraic closure of $K((X))$. Note that any injection from $B$ to itself can be extended to an endomorphism of $L$ over $K(X)$. There are many such injections, since $B$ is uncountable (see below for a proof). Restricting these endomorphisms to $K((X))$ gives many different homomorphisms $K((X))to L$ which all agree on $K[X]$. Restricting these homomorphisms to $K[[X]]$, we get many different homomorphisms $K[[X]]to L$ which agree on $K[X]$. (They are still different since a homomorphism on $K((X))$ is determined by its restriction to $K[[X]]$.)
[There are many other ways you could frame this same idea; for instance you could set it up more like uSir470888's answer. The point is that it's really easy to define homomorphisms into an algebraically closed field. In particular, once you're mapping into an algebraically closed field you can forget about whatever rigidity $K[[X]]$ might have had over $K[X]$ and map transcendental elements of $K[[X]]$ to any algebraically independent elements you want.]
Here is a proof that $K((X))$ always has uncountable transcendence degree over $K(X)$. First, we prove a lemma:
Lemma: Let $k$ be a field, let $K$ be a field extension of $k$, and suppose $Ssubset k((X))$ is algebraically independent over $k(X)$. Then $S$ is algebraically independent over $K(X)$, as a subset of $K((X))$.
Proof: Suppose some elements $s_1,dots,s_n in S$ satisfy $h(s_1,dots,s_n)=0$, where $h$ is a nonzero polynomial with coefficients in $K[X]$. For each $minmathbbZ$, the $X^m$ term of $h(s_1,dots,s_n)$ can be written as a polynomial in the coefficients of the $s_i$ (which are certain elements of $k$) and the coefficients of the coefficients of $h$ (each coefficient of $h$ is an element of $K[X]$, and the coefficients of the coefficients of $h$ are elements of $K$). Moreover, these polynomials are actually linear in the coefficients of the coefficients of $h$. So to say that $h(s_1,dots,s_n)=0$ is to say that the coefficients of the coefficients of $h$ satisfy a certain infinite list of linear equations with coefficients in $k$. But if a system of linear equations in finitely many variables with coefficients in a field $k$ has a nonzero solution in an extension of $k$, it has a nonzero solution in $k$ (this is essentially the fact that the rank of a matrix cannot change if you extend the base field). This means that we can replace the coefficients of $h$ by polynomials with coefficients in $k$ (which are not all zero) and $h(s_1,dots,s_n)=0$ will still be true. This contradicts the assumption that $S$ is algebraically independent over $k(X)$.
Now let $K$ be any field, and let $k$ be any countable subfield of $K$. Since $k(X)$ is countable and $k((X))$ is uncountable, there exists an uncountable subset $Ssubset k((X))$ which is algebraically independent over $k(X)$. By the Lemma, $S$ is also algebraically independent over $K(X)$ as a subset of $K((X))$.
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$begingroup$
Awesome! I wish I could accept several answers! The link stacks.math.columbia.edu/tag/00MC given by Mariano shows that $Atohat A_mathfrak m$ is not an epimorphism (unless it's an isomorphism) if $A$ is local and noetherian. I was wondering if, with your argument, you couldn't show that this is also true if $A$ is a (not necessarily noetherian) domain?
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– Pierre-Yves Gaillard
Aug 12 '17 at 17:02
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A variant of my argument shows that if $B$ is any $A$-algebra which has elements which are transcendental over $A$, then $Ato B$ is not an epimorphism. But I don't believe that $hatA_mathfrakm$ must always have transcendental elements (if it is different from $A$). For instance, let $A$ be local ring intermediate between $K[X]_(X)$ and $K[[X]]$ which contains an entire transcendence basis for $K[[X]]$. Then if I'm not mistaken, the completion of $A$ will just be $K[[X]]$, which is algebraic over $A$.
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– Eric Wofsey
Aug 12 '17 at 17:21
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Thank you! I realized that there is an elementary point I don't understand in you answer. You need to know (it seems to me) that the transcendence degree of $K((X))$ over $K(X)$ is at least two. What would be a simple argument to prove this?
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– Pierre-Yves Gaillard
Aug 12 '17 at 17:57
1
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I have added a proof that $K((X))$ has uncountable transcendence degree over $K(X)$ to the answer.
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– Eric Wofsey
Aug 12 '17 at 19:08
1
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Yeah, that's the structure of the argument.
$endgroup$
– Eric Wofsey
Aug 12 '17 at 19:42
|
show 2 more comments
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[Still checking if $g_1$ and $g_2$ are well defined. Let me know if you find something that doesn't. I changed the target ring to a larger one because I needed space for values on algebraic elements once I force a definition on a transcendental one.]
Let $R$ be the ring of power series on $X$ that are algebraic over the ring of fractions over $K$. This is $R$ consists of all power series $h$ such that there is a polynomial $p(Y)=a_0+a_1Y+...+a_nY^n$ with $a_i$ rational functions over $K$ with denominators that don't vanish at $X=0$, such that $p(h)=0$.
We need to construct two different homomorphisms $g_1,g_2$ from $K[[X]]$ to $R$ such that they coincide on the polynomials. That way $g_1circ i=g_2circ i$ for the inclusion $i:K[X]to K[[X]]$ but $g_1neq g_2$. Showing that $i$ is not an epimorphism.
Define $g_i(s)=s$ for $sin R subset K[[X]]$, $i=1,2$.
Now define $g_1(e^X)=1$ and $g_2(e^X)=sum_n=0^inftyX^n=(1-X)^-1$. In particular $g_1(e^X)=1neq (1-X)^-1=g_2(e^X)$. This defines $g_1,g_2$ on the sub-ring of $K[[X]]$ generated by $R$ and powers of $e^nX$ for $ninmathbbZ$, which are the elements of the form $sum_n=-N^Ne^nXa_n$ for $a_nin R$. Observe how the transcendence of $e^X$ implies that each element of this sub-ring has a unique expression in this form. Their images by $g_1,g_2$ are $sum_n=-N^Na_n$ and $sum_n=-N^N(1-X)^-na_n$ respectively.
We have $g_1,g_2$ defined in a larger sub-ring $R_1$. We still need to define them appropriately on the elements of $K[[X]]$ that are algebraic over $R_1$.
Assume $Yin K[[X]]$ is algebraic over $R_1$. Then there is a polynomial $p(Z)=a_0+a_1Z+...+a_nZ^n$ with coefficients $a_iin R_1$ such that $p(Y)=0$.
We need to define $g_1,g_2$ on $Y$ to be a solution of $0=g_i(a_0)+g_i(a_1)Z+...+g_i(a_n)Z^n$, respectively. In $R$ we have solutions of $0=g_i(a_0)+g_i(a_1)Z+...+g_i(a_n)Z^n$. So, we can send $Y$ to one of those solutions. This extends $g_1,g_2$ to a larger sub-ring $R_2$ of elements of $K[[X]]$ algebraic over $R_1$.
We can again pick an element of $K[[X]]$ transcendental over $R_2$. Defining their images in $R$ to be $1$ or $(1-X)^-1$ we can extend $g_1,g_2$ again and again to homomorphisms on ever larger subrings of $K[[X]]$.
The end results will be ring homomorphism that are different at $e^X$ but equal on the polynomials.
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Thanks! I'm trying to understand your answer, but I'm very slow. (I changed the index $K$ to $N$ in your post, to avoid a clash with the field $K$.)
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– Pierre-Yves Gaillard
Aug 12 '17 at 12:26
1
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Exp(-X) is not in the ring generated by R and exp(X), No?
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– Mariano Suárez-Álvarez
Aug 12 '17 at 14:04
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@MarianoSuárez-Álvarez Fixed that. I meant generated by the powers of $e^X$.
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– uSir470888
Aug 12 '17 at 14:26
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You can save yourself a lot of headaches by just taking your target to be some algebraically closed field containing $K[[X]]$; see my answer.
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– Eric Wofsey
Aug 12 '17 at 16:15
add a comment |
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3 Answers
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oldest
votes
3 Answers
3
active
oldest
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active
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active
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$begingroup$
The map from a Noetherian local ring to its completion is faithfully flat, and a faithfully flat epimorphism is an isomorphism. It follows from this that the map from the localization of $K[X]$ at $(X)$ to $K[[X]]$ is not epi.
$endgroup$
1
$begingroup$
See stacks.math.columbia.edu/tag/04VM and stacks.math.columbia.edu/tag/00MC
$endgroup$
– Mariano Suárez-Álvarez
Aug 12 '17 at 14:01
1
$begingroup$
Is there a straightforward way to extract a concrete counterexample from this?
$endgroup$
– tomasz
Aug 12 '17 at 14:08
1
$begingroup$
@Pierre-Yves Gaillard, we have that the inclusion factors as $K[X]to K[X]_(X)to K[[X]]$, so if the inclusion were epi, $K[X]_(X)to K[[X]]$ would have to be as well.
$endgroup$
– Ennar
Aug 12 '17 at 19:26
2
$begingroup$
I don't think that's what they're saying, they're saying "If $fcirc g$ is epi, then so is $f$"
$endgroup$
– Max
Aug 12 '17 at 20:03
1
$begingroup$
Yes, that's what I'm saying, @Max. That $K[X]to K[X]_(X)$ is epimorphism establishes the other direction in "if and only if", but is not needed here.
$endgroup$
– Ennar
Aug 12 '17 at 20:14
|
show 7 more comments
$begingroup$
The map from a Noetherian local ring to its completion is faithfully flat, and a faithfully flat epimorphism is an isomorphism. It follows from this that the map from the localization of $K[X]$ at $(X)$ to $K[[X]]$ is not epi.
$endgroup$
1
$begingroup$
See stacks.math.columbia.edu/tag/04VM and stacks.math.columbia.edu/tag/00MC
$endgroup$
– Mariano Suárez-Álvarez
Aug 12 '17 at 14:01
1
$begingroup$
Is there a straightforward way to extract a concrete counterexample from this?
$endgroup$
– tomasz
Aug 12 '17 at 14:08
1
$begingroup$
@Pierre-Yves Gaillard, we have that the inclusion factors as $K[X]to K[X]_(X)to K[[X]]$, so if the inclusion were epi, $K[X]_(X)to K[[X]]$ would have to be as well.
$endgroup$
– Ennar
Aug 12 '17 at 19:26
2
$begingroup$
I don't think that's what they're saying, they're saying "If $fcirc g$ is epi, then so is $f$"
$endgroup$
– Max
Aug 12 '17 at 20:03
1
$begingroup$
Yes, that's what I'm saying, @Max. That $K[X]to K[X]_(X)$ is epimorphism establishes the other direction in "if and only if", but is not needed here.
$endgroup$
– Ennar
Aug 12 '17 at 20:14
|
show 7 more comments
$begingroup$
The map from a Noetherian local ring to its completion is faithfully flat, and a faithfully flat epimorphism is an isomorphism. It follows from this that the map from the localization of $K[X]$ at $(X)$ to $K[[X]]$ is not epi.
$endgroup$
The map from a Noetherian local ring to its completion is faithfully flat, and a faithfully flat epimorphism is an isomorphism. It follows from this that the map from the localization of $K[X]$ at $(X)$ to $K[[X]]$ is not epi.
edited Jan 13 '18 at 1:36
Xam
4,57551846
4,57551846
answered Aug 12 '17 at 13:59
Mariano Suárez-ÁlvarezMariano Suárez-Álvarez
112k7157290
112k7157290
1
$begingroup$
See stacks.math.columbia.edu/tag/04VM and stacks.math.columbia.edu/tag/00MC
$endgroup$
– Mariano Suárez-Álvarez
Aug 12 '17 at 14:01
1
$begingroup$
Is there a straightforward way to extract a concrete counterexample from this?
$endgroup$
– tomasz
Aug 12 '17 at 14:08
1
$begingroup$
@Pierre-Yves Gaillard, we have that the inclusion factors as $K[X]to K[X]_(X)to K[[X]]$, so if the inclusion were epi, $K[X]_(X)to K[[X]]$ would have to be as well.
$endgroup$
– Ennar
Aug 12 '17 at 19:26
2
$begingroup$
I don't think that's what they're saying, they're saying "If $fcirc g$ is epi, then so is $f$"
$endgroup$
– Max
Aug 12 '17 at 20:03
1
$begingroup$
Yes, that's what I'm saying, @Max. That $K[X]to K[X]_(X)$ is epimorphism establishes the other direction in "if and only if", but is not needed here.
$endgroup$
– Ennar
Aug 12 '17 at 20:14
|
show 7 more comments
1
$begingroup$
See stacks.math.columbia.edu/tag/04VM and stacks.math.columbia.edu/tag/00MC
$endgroup$
– Mariano Suárez-Álvarez
Aug 12 '17 at 14:01
1
$begingroup$
Is there a straightforward way to extract a concrete counterexample from this?
$endgroup$
– tomasz
Aug 12 '17 at 14:08
1
$begingroup$
@Pierre-Yves Gaillard, we have that the inclusion factors as $K[X]to K[X]_(X)to K[[X]]$, so if the inclusion were epi, $K[X]_(X)to K[[X]]$ would have to be as well.
$endgroup$
– Ennar
Aug 12 '17 at 19:26
2
$begingroup$
I don't think that's what they're saying, they're saying "If $fcirc g$ is epi, then so is $f$"
$endgroup$
– Max
Aug 12 '17 at 20:03
1
$begingroup$
Yes, that's what I'm saying, @Max. That $K[X]to K[X]_(X)$ is epimorphism establishes the other direction in "if and only if", but is not needed here.
$endgroup$
– Ennar
Aug 12 '17 at 20:14
1
1
$begingroup$
See stacks.math.columbia.edu/tag/04VM and stacks.math.columbia.edu/tag/00MC
$endgroup$
– Mariano Suárez-Álvarez
Aug 12 '17 at 14:01
$begingroup$
See stacks.math.columbia.edu/tag/04VM and stacks.math.columbia.edu/tag/00MC
$endgroup$
– Mariano Suárez-Álvarez
Aug 12 '17 at 14:01
1
1
$begingroup$
Is there a straightforward way to extract a concrete counterexample from this?
$endgroup$
– tomasz
Aug 12 '17 at 14:08
$begingroup$
Is there a straightforward way to extract a concrete counterexample from this?
$endgroup$
– tomasz
Aug 12 '17 at 14:08
1
1
$begingroup$
@Pierre-Yves Gaillard, we have that the inclusion factors as $K[X]to K[X]_(X)to K[[X]]$, so if the inclusion were epi, $K[X]_(X)to K[[X]]$ would have to be as well.
$endgroup$
– Ennar
Aug 12 '17 at 19:26
$begingroup$
@Pierre-Yves Gaillard, we have that the inclusion factors as $K[X]to K[X]_(X)to K[[X]]$, so if the inclusion were epi, $K[X]_(X)to K[[X]]$ would have to be as well.
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– Ennar
Aug 12 '17 at 19:26
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I don't think that's what they're saying, they're saying "If $fcirc g$ is epi, then so is $f$"
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– Max
Aug 12 '17 at 20:03
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I don't think that's what they're saying, they're saying "If $fcirc g$ is epi, then so is $f$"
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– Max
Aug 12 '17 at 20:03
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Yes, that's what I'm saying, @Max. That $K[X]to K[X]_(X)$ is epimorphism establishes the other direction in "if and only if", but is not needed here.
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– Ennar
Aug 12 '17 at 20:14
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Yes, that's what I'm saying, @Max. That $K[X]to K[X]_(X)$ is epimorphism establishes the other direction in "if and only if", but is not needed here.
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– Ennar
Aug 12 '17 at 20:14
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show 7 more comments
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Let $B$ be a transcendence basis for the field $K((X))$ over $K(X)$ and let $L$ be an algebraic closure of $K((X))$. Note that any injection from $B$ to itself can be extended to an endomorphism of $L$ over $K(X)$. There are many such injections, since $B$ is uncountable (see below for a proof). Restricting these endomorphisms to $K((X))$ gives many different homomorphisms $K((X))to L$ which all agree on $K[X]$. Restricting these homomorphisms to $K[[X]]$, we get many different homomorphisms $K[[X]]to L$ which agree on $K[X]$. (They are still different since a homomorphism on $K((X))$ is determined by its restriction to $K[[X]]$.)
[There are many other ways you could frame this same idea; for instance you could set it up more like uSir470888's answer. The point is that it's really easy to define homomorphisms into an algebraically closed field. In particular, once you're mapping into an algebraically closed field you can forget about whatever rigidity $K[[X]]$ might have had over $K[X]$ and map transcendental elements of $K[[X]]$ to any algebraically independent elements you want.]
Here is a proof that $K((X))$ always has uncountable transcendence degree over $K(X)$. First, we prove a lemma:
Lemma: Let $k$ be a field, let $K$ be a field extension of $k$, and suppose $Ssubset k((X))$ is algebraically independent over $k(X)$. Then $S$ is algebraically independent over $K(X)$, as a subset of $K((X))$.
Proof: Suppose some elements $s_1,dots,s_n in S$ satisfy $h(s_1,dots,s_n)=0$, where $h$ is a nonzero polynomial with coefficients in $K[X]$. For each $minmathbbZ$, the $X^m$ term of $h(s_1,dots,s_n)$ can be written as a polynomial in the coefficients of the $s_i$ (which are certain elements of $k$) and the coefficients of the coefficients of $h$ (each coefficient of $h$ is an element of $K[X]$, and the coefficients of the coefficients of $h$ are elements of $K$). Moreover, these polynomials are actually linear in the coefficients of the coefficients of $h$. So to say that $h(s_1,dots,s_n)=0$ is to say that the coefficients of the coefficients of $h$ satisfy a certain infinite list of linear equations with coefficients in $k$. But if a system of linear equations in finitely many variables with coefficients in a field $k$ has a nonzero solution in an extension of $k$, it has a nonzero solution in $k$ (this is essentially the fact that the rank of a matrix cannot change if you extend the base field). This means that we can replace the coefficients of $h$ by polynomials with coefficients in $k$ (which are not all zero) and $h(s_1,dots,s_n)=0$ will still be true. This contradicts the assumption that $S$ is algebraically independent over $k(X)$.
Now let $K$ be any field, and let $k$ be any countable subfield of $K$. Since $k(X)$ is countable and $k((X))$ is uncountable, there exists an uncountable subset $Ssubset k((X))$ which is algebraically independent over $k(X)$. By the Lemma, $S$ is also algebraically independent over $K(X)$ as a subset of $K((X))$.
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Awesome! I wish I could accept several answers! The link stacks.math.columbia.edu/tag/00MC given by Mariano shows that $Atohat A_mathfrak m$ is not an epimorphism (unless it's an isomorphism) if $A$ is local and noetherian. I was wondering if, with your argument, you couldn't show that this is also true if $A$ is a (not necessarily noetherian) domain?
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– Pierre-Yves Gaillard
Aug 12 '17 at 17:02
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A variant of my argument shows that if $B$ is any $A$-algebra which has elements which are transcendental over $A$, then $Ato B$ is not an epimorphism. But I don't believe that $hatA_mathfrakm$ must always have transcendental elements (if it is different from $A$). For instance, let $A$ be local ring intermediate between $K[X]_(X)$ and $K[[X]]$ which contains an entire transcendence basis for $K[[X]]$. Then if I'm not mistaken, the completion of $A$ will just be $K[[X]]$, which is algebraic over $A$.
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– Eric Wofsey
Aug 12 '17 at 17:21
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Thank you! I realized that there is an elementary point I don't understand in you answer. You need to know (it seems to me) that the transcendence degree of $K((X))$ over $K(X)$ is at least two. What would be a simple argument to prove this?
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– Pierre-Yves Gaillard
Aug 12 '17 at 17:57
1
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I have added a proof that $K((X))$ has uncountable transcendence degree over $K(X)$ to the answer.
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– Eric Wofsey
Aug 12 '17 at 19:08
1
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Yeah, that's the structure of the argument.
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– Eric Wofsey
Aug 12 '17 at 19:42
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show 2 more comments
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Let $B$ be a transcendence basis for the field $K((X))$ over $K(X)$ and let $L$ be an algebraic closure of $K((X))$. Note that any injection from $B$ to itself can be extended to an endomorphism of $L$ over $K(X)$. There are many such injections, since $B$ is uncountable (see below for a proof). Restricting these endomorphisms to $K((X))$ gives many different homomorphisms $K((X))to L$ which all agree on $K[X]$. Restricting these homomorphisms to $K[[X]]$, we get many different homomorphisms $K[[X]]to L$ which agree on $K[X]$. (They are still different since a homomorphism on $K((X))$ is determined by its restriction to $K[[X]]$.)
[There are many other ways you could frame this same idea; for instance you could set it up more like uSir470888's answer. The point is that it's really easy to define homomorphisms into an algebraically closed field. In particular, once you're mapping into an algebraically closed field you can forget about whatever rigidity $K[[X]]$ might have had over $K[X]$ and map transcendental elements of $K[[X]]$ to any algebraically independent elements you want.]
Here is a proof that $K((X))$ always has uncountable transcendence degree over $K(X)$. First, we prove a lemma:
Lemma: Let $k$ be a field, let $K$ be a field extension of $k$, and suppose $Ssubset k((X))$ is algebraically independent over $k(X)$. Then $S$ is algebraically independent over $K(X)$, as a subset of $K((X))$.
Proof: Suppose some elements $s_1,dots,s_n in S$ satisfy $h(s_1,dots,s_n)=0$, where $h$ is a nonzero polynomial with coefficients in $K[X]$. For each $minmathbbZ$, the $X^m$ term of $h(s_1,dots,s_n)$ can be written as a polynomial in the coefficients of the $s_i$ (which are certain elements of $k$) and the coefficients of the coefficients of $h$ (each coefficient of $h$ is an element of $K[X]$, and the coefficients of the coefficients of $h$ are elements of $K$). Moreover, these polynomials are actually linear in the coefficients of the coefficients of $h$. So to say that $h(s_1,dots,s_n)=0$ is to say that the coefficients of the coefficients of $h$ satisfy a certain infinite list of linear equations with coefficients in $k$. But if a system of linear equations in finitely many variables with coefficients in a field $k$ has a nonzero solution in an extension of $k$, it has a nonzero solution in $k$ (this is essentially the fact that the rank of a matrix cannot change if you extend the base field). This means that we can replace the coefficients of $h$ by polynomials with coefficients in $k$ (which are not all zero) and $h(s_1,dots,s_n)=0$ will still be true. This contradicts the assumption that $S$ is algebraically independent over $k(X)$.
Now let $K$ be any field, and let $k$ be any countable subfield of $K$. Since $k(X)$ is countable and $k((X))$ is uncountable, there exists an uncountable subset $Ssubset k((X))$ which is algebraically independent over $k(X)$. By the Lemma, $S$ is also algebraically independent over $K(X)$ as a subset of $K((X))$.
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Awesome! I wish I could accept several answers! The link stacks.math.columbia.edu/tag/00MC given by Mariano shows that $Atohat A_mathfrak m$ is not an epimorphism (unless it's an isomorphism) if $A$ is local and noetherian. I was wondering if, with your argument, you couldn't show that this is also true if $A$ is a (not necessarily noetherian) domain?
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– Pierre-Yves Gaillard
Aug 12 '17 at 17:02
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A variant of my argument shows that if $B$ is any $A$-algebra which has elements which are transcendental over $A$, then $Ato B$ is not an epimorphism. But I don't believe that $hatA_mathfrakm$ must always have transcendental elements (if it is different from $A$). For instance, let $A$ be local ring intermediate between $K[X]_(X)$ and $K[[X]]$ which contains an entire transcendence basis for $K[[X]]$. Then if I'm not mistaken, the completion of $A$ will just be $K[[X]]$, which is algebraic over $A$.
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– Eric Wofsey
Aug 12 '17 at 17:21
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Thank you! I realized that there is an elementary point I don't understand in you answer. You need to know (it seems to me) that the transcendence degree of $K((X))$ over $K(X)$ is at least two. What would be a simple argument to prove this?
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– Pierre-Yves Gaillard
Aug 12 '17 at 17:57
1
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I have added a proof that $K((X))$ has uncountable transcendence degree over $K(X)$ to the answer.
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– Eric Wofsey
Aug 12 '17 at 19:08
1
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Yeah, that's the structure of the argument.
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– Eric Wofsey
Aug 12 '17 at 19:42
|
show 2 more comments
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Let $B$ be a transcendence basis for the field $K((X))$ over $K(X)$ and let $L$ be an algebraic closure of $K((X))$. Note that any injection from $B$ to itself can be extended to an endomorphism of $L$ over $K(X)$. There are many such injections, since $B$ is uncountable (see below for a proof). Restricting these endomorphisms to $K((X))$ gives many different homomorphisms $K((X))to L$ which all agree on $K[X]$. Restricting these homomorphisms to $K[[X]]$, we get many different homomorphisms $K[[X]]to L$ which agree on $K[X]$. (They are still different since a homomorphism on $K((X))$ is determined by its restriction to $K[[X]]$.)
[There are many other ways you could frame this same idea; for instance you could set it up more like uSir470888's answer. The point is that it's really easy to define homomorphisms into an algebraically closed field. In particular, once you're mapping into an algebraically closed field you can forget about whatever rigidity $K[[X]]$ might have had over $K[X]$ and map transcendental elements of $K[[X]]$ to any algebraically independent elements you want.]
Here is a proof that $K((X))$ always has uncountable transcendence degree over $K(X)$. First, we prove a lemma:
Lemma: Let $k$ be a field, let $K$ be a field extension of $k$, and suppose $Ssubset k((X))$ is algebraically independent over $k(X)$. Then $S$ is algebraically independent over $K(X)$, as a subset of $K((X))$.
Proof: Suppose some elements $s_1,dots,s_n in S$ satisfy $h(s_1,dots,s_n)=0$, where $h$ is a nonzero polynomial with coefficients in $K[X]$. For each $minmathbbZ$, the $X^m$ term of $h(s_1,dots,s_n)$ can be written as a polynomial in the coefficients of the $s_i$ (which are certain elements of $k$) and the coefficients of the coefficients of $h$ (each coefficient of $h$ is an element of $K[X]$, and the coefficients of the coefficients of $h$ are elements of $K$). Moreover, these polynomials are actually linear in the coefficients of the coefficients of $h$. So to say that $h(s_1,dots,s_n)=0$ is to say that the coefficients of the coefficients of $h$ satisfy a certain infinite list of linear equations with coefficients in $k$. But if a system of linear equations in finitely many variables with coefficients in a field $k$ has a nonzero solution in an extension of $k$, it has a nonzero solution in $k$ (this is essentially the fact that the rank of a matrix cannot change if you extend the base field). This means that we can replace the coefficients of $h$ by polynomials with coefficients in $k$ (which are not all zero) and $h(s_1,dots,s_n)=0$ will still be true. This contradicts the assumption that $S$ is algebraically independent over $k(X)$.
Now let $K$ be any field, and let $k$ be any countable subfield of $K$. Since $k(X)$ is countable and $k((X))$ is uncountable, there exists an uncountable subset $Ssubset k((X))$ which is algebraically independent over $k(X)$. By the Lemma, $S$ is also algebraically independent over $K(X)$ as a subset of $K((X))$.
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Let $B$ be a transcendence basis for the field $K((X))$ over $K(X)$ and let $L$ be an algebraic closure of $K((X))$. Note that any injection from $B$ to itself can be extended to an endomorphism of $L$ over $K(X)$. There are many such injections, since $B$ is uncountable (see below for a proof). Restricting these endomorphisms to $K((X))$ gives many different homomorphisms $K((X))to L$ which all agree on $K[X]$. Restricting these homomorphisms to $K[[X]]$, we get many different homomorphisms $K[[X]]to L$ which agree on $K[X]$. (They are still different since a homomorphism on $K((X))$ is determined by its restriction to $K[[X]]$.)
[There are many other ways you could frame this same idea; for instance you could set it up more like uSir470888's answer. The point is that it's really easy to define homomorphisms into an algebraically closed field. In particular, once you're mapping into an algebraically closed field you can forget about whatever rigidity $K[[X]]$ might have had over $K[X]$ and map transcendental elements of $K[[X]]$ to any algebraically independent elements you want.]
Here is a proof that $K((X))$ always has uncountable transcendence degree over $K(X)$. First, we prove a lemma:
Lemma: Let $k$ be a field, let $K$ be a field extension of $k$, and suppose $Ssubset k((X))$ is algebraically independent over $k(X)$. Then $S$ is algebraically independent over $K(X)$, as a subset of $K((X))$.
Proof: Suppose some elements $s_1,dots,s_n in S$ satisfy $h(s_1,dots,s_n)=0$, where $h$ is a nonzero polynomial with coefficients in $K[X]$. For each $minmathbbZ$, the $X^m$ term of $h(s_1,dots,s_n)$ can be written as a polynomial in the coefficients of the $s_i$ (which are certain elements of $k$) and the coefficients of the coefficients of $h$ (each coefficient of $h$ is an element of $K[X]$, and the coefficients of the coefficients of $h$ are elements of $K$). Moreover, these polynomials are actually linear in the coefficients of the coefficients of $h$. So to say that $h(s_1,dots,s_n)=0$ is to say that the coefficients of the coefficients of $h$ satisfy a certain infinite list of linear equations with coefficients in $k$. But if a system of linear equations in finitely many variables with coefficients in a field $k$ has a nonzero solution in an extension of $k$, it has a nonzero solution in $k$ (this is essentially the fact that the rank of a matrix cannot change if you extend the base field). This means that we can replace the coefficients of $h$ by polynomials with coefficients in $k$ (which are not all zero) and $h(s_1,dots,s_n)=0$ will still be true. This contradicts the assumption that $S$ is algebraically independent over $k(X)$.
Now let $K$ be any field, and let $k$ be any countable subfield of $K$. Since $k(X)$ is countable and $k((X))$ is uncountable, there exists an uncountable subset $Ssubset k((X))$ which is algebraically independent over $k(X)$. By the Lemma, $S$ is also algebraically independent over $K(X)$ as a subset of $K((X))$.
edited Mar 17 at 23:08
answered Aug 12 '17 at 16:10
Eric WofseyEric Wofsey
191k14216349
191k14216349
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Awesome! I wish I could accept several answers! The link stacks.math.columbia.edu/tag/00MC given by Mariano shows that $Atohat A_mathfrak m$ is not an epimorphism (unless it's an isomorphism) if $A$ is local and noetherian. I was wondering if, with your argument, you couldn't show that this is also true if $A$ is a (not necessarily noetherian) domain?
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– Pierre-Yves Gaillard
Aug 12 '17 at 17:02
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A variant of my argument shows that if $B$ is any $A$-algebra which has elements which are transcendental over $A$, then $Ato B$ is not an epimorphism. But I don't believe that $hatA_mathfrakm$ must always have transcendental elements (if it is different from $A$). For instance, let $A$ be local ring intermediate between $K[X]_(X)$ and $K[[X]]$ which contains an entire transcendence basis for $K[[X]]$. Then if I'm not mistaken, the completion of $A$ will just be $K[[X]]$, which is algebraic over $A$.
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– Eric Wofsey
Aug 12 '17 at 17:21
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Thank you! I realized that there is an elementary point I don't understand in you answer. You need to know (it seems to me) that the transcendence degree of $K((X))$ over $K(X)$ is at least two. What would be a simple argument to prove this?
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– Pierre-Yves Gaillard
Aug 12 '17 at 17:57
1
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I have added a proof that $K((X))$ has uncountable transcendence degree over $K(X)$ to the answer.
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– Eric Wofsey
Aug 12 '17 at 19:08
1
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Yeah, that's the structure of the argument.
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– Eric Wofsey
Aug 12 '17 at 19:42
|
show 2 more comments
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Awesome! I wish I could accept several answers! The link stacks.math.columbia.edu/tag/00MC given by Mariano shows that $Atohat A_mathfrak m$ is not an epimorphism (unless it's an isomorphism) if $A$ is local and noetherian. I was wondering if, with your argument, you couldn't show that this is also true if $A$ is a (not necessarily noetherian) domain?
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– Pierre-Yves Gaillard
Aug 12 '17 at 17:02
$begingroup$
A variant of my argument shows that if $B$ is any $A$-algebra which has elements which are transcendental over $A$, then $Ato B$ is not an epimorphism. But I don't believe that $hatA_mathfrakm$ must always have transcendental elements (if it is different from $A$). For instance, let $A$ be local ring intermediate between $K[X]_(X)$ and $K[[X]]$ which contains an entire transcendence basis for $K[[X]]$. Then if I'm not mistaken, the completion of $A$ will just be $K[[X]]$, which is algebraic over $A$.
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– Eric Wofsey
Aug 12 '17 at 17:21
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Thank you! I realized that there is an elementary point I don't understand in you answer. You need to know (it seems to me) that the transcendence degree of $K((X))$ over $K(X)$ is at least two. What would be a simple argument to prove this?
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– Pierre-Yves Gaillard
Aug 12 '17 at 17:57
1
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I have added a proof that $K((X))$ has uncountable transcendence degree over $K(X)$ to the answer.
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– Eric Wofsey
Aug 12 '17 at 19:08
1
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Yeah, that's the structure of the argument.
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– Eric Wofsey
Aug 12 '17 at 19:42
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Awesome! I wish I could accept several answers! The link stacks.math.columbia.edu/tag/00MC given by Mariano shows that $Atohat A_mathfrak m$ is not an epimorphism (unless it's an isomorphism) if $A$ is local and noetherian. I was wondering if, with your argument, you couldn't show that this is also true if $A$ is a (not necessarily noetherian) domain?
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– Pierre-Yves Gaillard
Aug 12 '17 at 17:02
$begingroup$
Awesome! I wish I could accept several answers! The link stacks.math.columbia.edu/tag/00MC given by Mariano shows that $Atohat A_mathfrak m$ is not an epimorphism (unless it's an isomorphism) if $A$ is local and noetherian. I was wondering if, with your argument, you couldn't show that this is also true if $A$ is a (not necessarily noetherian) domain?
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– Pierre-Yves Gaillard
Aug 12 '17 at 17:02
$begingroup$
A variant of my argument shows that if $B$ is any $A$-algebra which has elements which are transcendental over $A$, then $Ato B$ is not an epimorphism. But I don't believe that $hatA_mathfrakm$ must always have transcendental elements (if it is different from $A$). For instance, let $A$ be local ring intermediate between $K[X]_(X)$ and $K[[X]]$ which contains an entire transcendence basis for $K[[X]]$. Then if I'm not mistaken, the completion of $A$ will just be $K[[X]]$, which is algebraic over $A$.
$endgroup$
– Eric Wofsey
Aug 12 '17 at 17:21
$begingroup$
A variant of my argument shows that if $B$ is any $A$-algebra which has elements which are transcendental over $A$, then $Ato B$ is not an epimorphism. But I don't believe that $hatA_mathfrakm$ must always have transcendental elements (if it is different from $A$). For instance, let $A$ be local ring intermediate between $K[X]_(X)$ and $K[[X]]$ which contains an entire transcendence basis for $K[[X]]$. Then if I'm not mistaken, the completion of $A$ will just be $K[[X]]$, which is algebraic over $A$.
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– Eric Wofsey
Aug 12 '17 at 17:21
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Thank you! I realized that there is an elementary point I don't understand in you answer. You need to know (it seems to me) that the transcendence degree of $K((X))$ over $K(X)$ is at least two. What would be a simple argument to prove this?
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– Pierre-Yves Gaillard
Aug 12 '17 at 17:57
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Thank you! I realized that there is an elementary point I don't understand in you answer. You need to know (it seems to me) that the transcendence degree of $K((X))$ over $K(X)$ is at least two. What would be a simple argument to prove this?
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– Pierre-Yves Gaillard
Aug 12 '17 at 17:57
1
1
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I have added a proof that $K((X))$ has uncountable transcendence degree over $K(X)$ to the answer.
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– Eric Wofsey
Aug 12 '17 at 19:08
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I have added a proof that $K((X))$ has uncountable transcendence degree over $K(X)$ to the answer.
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– Eric Wofsey
Aug 12 '17 at 19:08
1
1
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Yeah, that's the structure of the argument.
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– Eric Wofsey
Aug 12 '17 at 19:42
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Yeah, that's the structure of the argument.
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– Eric Wofsey
Aug 12 '17 at 19:42
|
show 2 more comments
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[Still checking if $g_1$ and $g_2$ are well defined. Let me know if you find something that doesn't. I changed the target ring to a larger one because I needed space for values on algebraic elements once I force a definition on a transcendental one.]
Let $R$ be the ring of power series on $X$ that are algebraic over the ring of fractions over $K$. This is $R$ consists of all power series $h$ such that there is a polynomial $p(Y)=a_0+a_1Y+...+a_nY^n$ with $a_i$ rational functions over $K$ with denominators that don't vanish at $X=0$, such that $p(h)=0$.
We need to construct two different homomorphisms $g_1,g_2$ from $K[[X]]$ to $R$ such that they coincide on the polynomials. That way $g_1circ i=g_2circ i$ for the inclusion $i:K[X]to K[[X]]$ but $g_1neq g_2$. Showing that $i$ is not an epimorphism.
Define $g_i(s)=s$ for $sin R subset K[[X]]$, $i=1,2$.
Now define $g_1(e^X)=1$ and $g_2(e^X)=sum_n=0^inftyX^n=(1-X)^-1$. In particular $g_1(e^X)=1neq (1-X)^-1=g_2(e^X)$. This defines $g_1,g_2$ on the sub-ring of $K[[X]]$ generated by $R$ and powers of $e^nX$ for $ninmathbbZ$, which are the elements of the form $sum_n=-N^Ne^nXa_n$ for $a_nin R$. Observe how the transcendence of $e^X$ implies that each element of this sub-ring has a unique expression in this form. Their images by $g_1,g_2$ are $sum_n=-N^Na_n$ and $sum_n=-N^N(1-X)^-na_n$ respectively.
We have $g_1,g_2$ defined in a larger sub-ring $R_1$. We still need to define them appropriately on the elements of $K[[X]]$ that are algebraic over $R_1$.
Assume $Yin K[[X]]$ is algebraic over $R_1$. Then there is a polynomial $p(Z)=a_0+a_1Z+...+a_nZ^n$ with coefficients $a_iin R_1$ such that $p(Y)=0$.
We need to define $g_1,g_2$ on $Y$ to be a solution of $0=g_i(a_0)+g_i(a_1)Z+...+g_i(a_n)Z^n$, respectively. In $R$ we have solutions of $0=g_i(a_0)+g_i(a_1)Z+...+g_i(a_n)Z^n$. So, we can send $Y$ to one of those solutions. This extends $g_1,g_2$ to a larger sub-ring $R_2$ of elements of $K[[X]]$ algebraic over $R_1$.
We can again pick an element of $K[[X]]$ transcendental over $R_2$. Defining their images in $R$ to be $1$ or $(1-X)^-1$ we can extend $g_1,g_2$ again and again to homomorphisms on ever larger subrings of $K[[X]]$.
The end results will be ring homomorphism that are different at $e^X$ but equal on the polynomials.
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Thanks! I'm trying to understand your answer, but I'm very slow. (I changed the index $K$ to $N$ in your post, to avoid a clash with the field $K$.)
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– Pierre-Yves Gaillard
Aug 12 '17 at 12:26
1
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Exp(-X) is not in the ring generated by R and exp(X), No?
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– Mariano Suárez-Álvarez
Aug 12 '17 at 14:04
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@MarianoSuárez-Álvarez Fixed that. I meant generated by the powers of $e^X$.
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– uSir470888
Aug 12 '17 at 14:26
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You can save yourself a lot of headaches by just taking your target to be some algebraically closed field containing $K[[X]]$; see my answer.
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– Eric Wofsey
Aug 12 '17 at 16:15
add a comment |
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[Still checking if $g_1$ and $g_2$ are well defined. Let me know if you find something that doesn't. I changed the target ring to a larger one because I needed space for values on algebraic elements once I force a definition on a transcendental one.]
Let $R$ be the ring of power series on $X$ that are algebraic over the ring of fractions over $K$. This is $R$ consists of all power series $h$ such that there is a polynomial $p(Y)=a_0+a_1Y+...+a_nY^n$ with $a_i$ rational functions over $K$ with denominators that don't vanish at $X=0$, such that $p(h)=0$.
We need to construct two different homomorphisms $g_1,g_2$ from $K[[X]]$ to $R$ such that they coincide on the polynomials. That way $g_1circ i=g_2circ i$ for the inclusion $i:K[X]to K[[X]]$ but $g_1neq g_2$. Showing that $i$ is not an epimorphism.
Define $g_i(s)=s$ for $sin R subset K[[X]]$, $i=1,2$.
Now define $g_1(e^X)=1$ and $g_2(e^X)=sum_n=0^inftyX^n=(1-X)^-1$. In particular $g_1(e^X)=1neq (1-X)^-1=g_2(e^X)$. This defines $g_1,g_2$ on the sub-ring of $K[[X]]$ generated by $R$ and powers of $e^nX$ for $ninmathbbZ$, which are the elements of the form $sum_n=-N^Ne^nXa_n$ for $a_nin R$. Observe how the transcendence of $e^X$ implies that each element of this sub-ring has a unique expression in this form. Their images by $g_1,g_2$ are $sum_n=-N^Na_n$ and $sum_n=-N^N(1-X)^-na_n$ respectively.
We have $g_1,g_2$ defined in a larger sub-ring $R_1$. We still need to define them appropriately on the elements of $K[[X]]$ that are algebraic over $R_1$.
Assume $Yin K[[X]]$ is algebraic over $R_1$. Then there is a polynomial $p(Z)=a_0+a_1Z+...+a_nZ^n$ with coefficients $a_iin R_1$ such that $p(Y)=0$.
We need to define $g_1,g_2$ on $Y$ to be a solution of $0=g_i(a_0)+g_i(a_1)Z+...+g_i(a_n)Z^n$, respectively. In $R$ we have solutions of $0=g_i(a_0)+g_i(a_1)Z+...+g_i(a_n)Z^n$. So, we can send $Y$ to one of those solutions. This extends $g_1,g_2$ to a larger sub-ring $R_2$ of elements of $K[[X]]$ algebraic over $R_1$.
We can again pick an element of $K[[X]]$ transcendental over $R_2$. Defining their images in $R$ to be $1$ or $(1-X)^-1$ we can extend $g_1,g_2$ again and again to homomorphisms on ever larger subrings of $K[[X]]$.
The end results will be ring homomorphism that are different at $e^X$ but equal on the polynomials.
$endgroup$
$begingroup$
Thanks! I'm trying to understand your answer, but I'm very slow. (I changed the index $K$ to $N$ in your post, to avoid a clash with the field $K$.)
$endgroup$
– Pierre-Yves Gaillard
Aug 12 '17 at 12:26
1
$begingroup$
Exp(-X) is not in the ring generated by R and exp(X), No?
$endgroup$
– Mariano Suárez-Álvarez
Aug 12 '17 at 14:04
$begingroup$
@MarianoSuárez-Álvarez Fixed that. I meant generated by the powers of $e^X$.
$endgroup$
– uSir470888
Aug 12 '17 at 14:26
$begingroup$
You can save yourself a lot of headaches by just taking your target to be some algebraically closed field containing $K[[X]]$; see my answer.
$endgroup$
– Eric Wofsey
Aug 12 '17 at 16:15
add a comment |
$begingroup$
[Still checking if $g_1$ and $g_2$ are well defined. Let me know if you find something that doesn't. I changed the target ring to a larger one because I needed space for values on algebraic elements once I force a definition on a transcendental one.]
Let $R$ be the ring of power series on $X$ that are algebraic over the ring of fractions over $K$. This is $R$ consists of all power series $h$ such that there is a polynomial $p(Y)=a_0+a_1Y+...+a_nY^n$ with $a_i$ rational functions over $K$ with denominators that don't vanish at $X=0$, such that $p(h)=0$.
We need to construct two different homomorphisms $g_1,g_2$ from $K[[X]]$ to $R$ such that they coincide on the polynomials. That way $g_1circ i=g_2circ i$ for the inclusion $i:K[X]to K[[X]]$ but $g_1neq g_2$. Showing that $i$ is not an epimorphism.
Define $g_i(s)=s$ for $sin R subset K[[X]]$, $i=1,2$.
Now define $g_1(e^X)=1$ and $g_2(e^X)=sum_n=0^inftyX^n=(1-X)^-1$. In particular $g_1(e^X)=1neq (1-X)^-1=g_2(e^X)$. This defines $g_1,g_2$ on the sub-ring of $K[[X]]$ generated by $R$ and powers of $e^nX$ for $ninmathbbZ$, which are the elements of the form $sum_n=-N^Ne^nXa_n$ for $a_nin R$. Observe how the transcendence of $e^X$ implies that each element of this sub-ring has a unique expression in this form. Their images by $g_1,g_2$ are $sum_n=-N^Na_n$ and $sum_n=-N^N(1-X)^-na_n$ respectively.
We have $g_1,g_2$ defined in a larger sub-ring $R_1$. We still need to define them appropriately on the elements of $K[[X]]$ that are algebraic over $R_1$.
Assume $Yin K[[X]]$ is algebraic over $R_1$. Then there is a polynomial $p(Z)=a_0+a_1Z+...+a_nZ^n$ with coefficients $a_iin R_1$ such that $p(Y)=0$.
We need to define $g_1,g_2$ on $Y$ to be a solution of $0=g_i(a_0)+g_i(a_1)Z+...+g_i(a_n)Z^n$, respectively. In $R$ we have solutions of $0=g_i(a_0)+g_i(a_1)Z+...+g_i(a_n)Z^n$. So, we can send $Y$ to one of those solutions. This extends $g_1,g_2$ to a larger sub-ring $R_2$ of elements of $K[[X]]$ algebraic over $R_1$.
We can again pick an element of $K[[X]]$ transcendental over $R_2$. Defining their images in $R$ to be $1$ or $(1-X)^-1$ we can extend $g_1,g_2$ again and again to homomorphisms on ever larger subrings of $K[[X]]$.
The end results will be ring homomorphism that are different at $e^X$ but equal on the polynomials.
$endgroup$
[Still checking if $g_1$ and $g_2$ are well defined. Let me know if you find something that doesn't. I changed the target ring to a larger one because I needed space for values on algebraic elements once I force a definition on a transcendental one.]
Let $R$ be the ring of power series on $X$ that are algebraic over the ring of fractions over $K$. This is $R$ consists of all power series $h$ such that there is a polynomial $p(Y)=a_0+a_1Y+...+a_nY^n$ with $a_i$ rational functions over $K$ with denominators that don't vanish at $X=0$, such that $p(h)=0$.
We need to construct two different homomorphisms $g_1,g_2$ from $K[[X]]$ to $R$ such that they coincide on the polynomials. That way $g_1circ i=g_2circ i$ for the inclusion $i:K[X]to K[[X]]$ but $g_1neq g_2$. Showing that $i$ is not an epimorphism.
Define $g_i(s)=s$ for $sin R subset K[[X]]$, $i=1,2$.
Now define $g_1(e^X)=1$ and $g_2(e^X)=sum_n=0^inftyX^n=(1-X)^-1$. In particular $g_1(e^X)=1neq (1-X)^-1=g_2(e^X)$. This defines $g_1,g_2$ on the sub-ring of $K[[X]]$ generated by $R$ and powers of $e^nX$ for $ninmathbbZ$, which are the elements of the form $sum_n=-N^Ne^nXa_n$ for $a_nin R$. Observe how the transcendence of $e^X$ implies that each element of this sub-ring has a unique expression in this form. Their images by $g_1,g_2$ are $sum_n=-N^Na_n$ and $sum_n=-N^N(1-X)^-na_n$ respectively.
We have $g_1,g_2$ defined in a larger sub-ring $R_1$. We still need to define them appropriately on the elements of $K[[X]]$ that are algebraic over $R_1$.
Assume $Yin K[[X]]$ is algebraic over $R_1$. Then there is a polynomial $p(Z)=a_0+a_1Z+...+a_nZ^n$ with coefficients $a_iin R_1$ such that $p(Y)=0$.
We need to define $g_1,g_2$ on $Y$ to be a solution of $0=g_i(a_0)+g_i(a_1)Z+...+g_i(a_n)Z^n$, respectively. In $R$ we have solutions of $0=g_i(a_0)+g_i(a_1)Z+...+g_i(a_n)Z^n$. So, we can send $Y$ to one of those solutions. This extends $g_1,g_2$ to a larger sub-ring $R_2$ of elements of $K[[X]]$ algebraic over $R_1$.
We can again pick an element of $K[[X]]$ transcendental over $R_2$. Defining their images in $R$ to be $1$ or $(1-X)^-1$ we can extend $g_1,g_2$ again and again to homomorphisms on ever larger subrings of $K[[X]]$.
The end results will be ring homomorphism that are different at $e^X$ but equal on the polynomials.
edited Aug 12 '17 at 14:10
answered Aug 12 '17 at 12:11
uSir470888uSir470888
27026
27026
$begingroup$
Thanks! I'm trying to understand your answer, but I'm very slow. (I changed the index $K$ to $N$ in your post, to avoid a clash with the field $K$.)
$endgroup$
– Pierre-Yves Gaillard
Aug 12 '17 at 12:26
1
$begingroup$
Exp(-X) is not in the ring generated by R and exp(X), No?
$endgroup$
– Mariano Suárez-Álvarez
Aug 12 '17 at 14:04
$begingroup$
@MarianoSuárez-Álvarez Fixed that. I meant generated by the powers of $e^X$.
$endgroup$
– uSir470888
Aug 12 '17 at 14:26
$begingroup$
You can save yourself a lot of headaches by just taking your target to be some algebraically closed field containing $K[[X]]$; see my answer.
$endgroup$
– Eric Wofsey
Aug 12 '17 at 16:15
add a comment |
$begingroup$
Thanks! I'm trying to understand your answer, but I'm very slow. (I changed the index $K$ to $N$ in your post, to avoid a clash with the field $K$.)
$endgroup$
– Pierre-Yves Gaillard
Aug 12 '17 at 12:26
1
$begingroup$
Exp(-X) is not in the ring generated by R and exp(X), No?
$endgroup$
– Mariano Suárez-Álvarez
Aug 12 '17 at 14:04
$begingroup$
@MarianoSuárez-Álvarez Fixed that. I meant generated by the powers of $e^X$.
$endgroup$
– uSir470888
Aug 12 '17 at 14:26
$begingroup$
You can save yourself a lot of headaches by just taking your target to be some algebraically closed field containing $K[[X]]$; see my answer.
$endgroup$
– Eric Wofsey
Aug 12 '17 at 16:15
$begingroup$
Thanks! I'm trying to understand your answer, but I'm very slow. (I changed the index $K$ to $N$ in your post, to avoid a clash with the field $K$.)
$endgroup$
– Pierre-Yves Gaillard
Aug 12 '17 at 12:26
$begingroup$
Thanks! I'm trying to understand your answer, but I'm very slow. (I changed the index $K$ to $N$ in your post, to avoid a clash with the field $K$.)
$endgroup$
– Pierre-Yves Gaillard
Aug 12 '17 at 12:26
1
1
$begingroup$
Exp(-X) is not in the ring generated by R and exp(X), No?
$endgroup$
– Mariano Suárez-Álvarez
Aug 12 '17 at 14:04
$begingroup$
Exp(-X) is not in the ring generated by R and exp(X), No?
$endgroup$
– Mariano Suárez-Álvarez
Aug 12 '17 at 14:04
$begingroup$
@MarianoSuárez-Álvarez Fixed that. I meant generated by the powers of $e^X$.
$endgroup$
– uSir470888
Aug 12 '17 at 14:26
$begingroup$
@MarianoSuárez-Álvarez Fixed that. I meant generated by the powers of $e^X$.
$endgroup$
– uSir470888
Aug 12 '17 at 14:26
$begingroup$
You can save yourself a lot of headaches by just taking your target to be some algebraically closed field containing $K[[X]]$; see my answer.
$endgroup$
– Eric Wofsey
Aug 12 '17 at 16:15
$begingroup$
You can save yourself a lot of headaches by just taking your target to be some algebraically closed field containing $K[[X]]$; see my answer.
$endgroup$
– Eric Wofsey
Aug 12 '17 at 16:15
add a comment |
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$begingroup$
What is an epimorphism exactly ? $K[[x]]$ is the completion of $K[x]$ for the (non-Archimedian) absolute value $|x^Nf(x)| = 2^-N$ if $f(0) ne 0$
$endgroup$
– reuns
Aug 12 '17 at 12:56
2
$begingroup$
@reuns The inclusion $f:K[X]to K[[X]]$ is epi if for every two ring homomorphisms $g_1,g_2:K[[X]]to R$, if $g_1circ f=g_2circ f$ then $g_1=g_2$.
$endgroup$
– uSir470888
Aug 12 '17 at 13:00